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  • 2019中山大学程序设计竞赛 Enlarge it(水题)2019-04-20 16:51:05

    Problem Description Taotao is glad to see that lots of students have registered for today's algorithm competition. Thus he wants to pull a banner on the wall. But Taotao is so stupid that he doesn't even know how to enlarge words to fit the size

  • 1002. Find Common Characters2019-04-13 11:41:10

    Given an array A of strings made only from lowercase letters, return a list of all characters that show up in all strings within the list (including duplicates).  For example, if a character occurs 3 times in all strings but not 4 times, you need to inc

  • JavaScript String常用方法和属性2019-04-12 22:53:37

     一、string中的常用属性    1. length 说明:   字符串的长度属性,一直保持跟踪着该字符串中包含的字符数。      二、string中的常用方法    1. toLowerCase() 和 toUpperCase() 说明:   toLowerCase(): returns a copy of the string with its letters converted to low

  • LeetCode算法题-1-bit and 2-bit Characters(Java实现)2019-04-10 08:51:49

    这是悦乐书的第302次更新,第321篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第170题(顺位题号是717)。有两个特殊字符,第一个字符可以用一个比特0表示,第二个字符可以用两个比特(10或11)表示。现在给出一个由比特位组成的数组,判断其最后一个字符是否是一位字符。数组的

  • Python编程之求字符串长度2019-04-07 09:50:45

    问题描述:写一个函数,求一个字符串的长度,在main函数中输入字符串,并输出其长度。 源代码: #!/usr/bin/python # -*- coding: UTF-8 -*- if __name__ == '__main__': s = raw_input('please input a string:\n') print 'the string has %d characters.' % len(s) 输出结

  • CSI2120 Programming Paradigms2019-03-31 19:47:36

    Assignment 3CSI2120 Programming ParadigmsWinter 2019Due on April 5th before 11:00 pm in Virtual Campus6 marksThere are [10 points] in this assignment. The assignment is worth 6% of your final mark.All code must be submitted in a scm file. Screenshots, fil

  • 999. Available Captures for Rook2019-03-20 10:54:22

    On an 8 x 8 chessboard, there is one white rook. There also may be empty squares, white bishops, and black pawns. These are given as characters ‘R’, ‘.’, ‘B’, and ‘p’ respectively. Uppercase characters represent white pieces, and lowercase charac

  • c/c++的常用函数和STL使用2019-03-07 23:43:33

    一个超好用的c++网站:http://www.cplusplus.com/reference/string/string/erase/ 一、string类的函数 1.string& erase (size_t pos = 0, size_t len = npos); pos: Position of the first character to be erased.If this is greater than the string length, it throws out_of_ra

  • #Leetcode# 541. Reverse String II2019-02-13 21:54:28

    https://leetcode.com/problems/reverse-string-ii/   Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of them. If

  • 报错:The valid characters are defined in RFC 7230 and RFC 39862019-02-11 13:50:24

      访问 spring boot controller时,报错:The valid characters are defined in RFC 7230 and RFC 3986 解决: @SpringBootApplication public class Application { public static void main(String[] args) { SpringApplication.run(Application.class, args);

  • Leetcode Summary...2019-02-09 18:37:47

    Sliding Window Problems 76. Minimum Window Substring 438. Find All Anagrams in a String 30. Substring with Concatenation of All Words 3. Longest Substring Without Repeating Characters 159. Longest Substring with At Most Two Distinct Characters 567. Permut

  • [leetcode] reorganize String2019-02-06 19:40:27

    Runtime: 4 ms; Memory Usage: 5 MB (好高啊魂淡!again   我懒死算了这样下去欠了多少债 //大致是参考solution1的思路 bool compare(pair<int,char>a, pair<int,char>b) { return a.first>b.first; } class Solution { public: string reorganizeString(string S) {

  • #Leetcode# 451. Sort Characters By Frequency2019-01-30 20:42:38

    https://leetcode.com/problems/sort-characters-by-frequency/   Given a string, sort it in decreasing order based on the frequency of characters. Example 1: Input: "tree" Output: "eert" Explanation: 'e' appears twice while &#

  • leetcode 344. Reverse String2019-01-30 20:39:00

    Write a function that reverses a string. The input string is given as an array of characters char[]. Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory. You may assume all the chara

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